Galitski's Exploring Quantum Mechanics says (on p.29) that

the number of (linearly) independent unitary ($N$-dimensional) matrices is also $N^2$.

Since the set of unitary matrices does not form a vector space, I was curious about how to construct such a unitary basis of $F^{n\times n}$. So I ended up reading Nicholas Wheeler's note on unitary basis, where he refers to the Weyl's unitary basis; $$E(\alpha,\beta) = e^{{i\over \hbar}({\alpha p + \beta x})}$$ with the property $${1\over h}\text{Tr}(E(\alpha,\beta))=\delta(\alpha)\delta(\beta)$$ But I don't get

  1. How is the property derived from the BCH formula?

    My attempt : $$\text{Tr}(E(\alpha,\beta)) = \int dx\left<x\right|e^{{i\over \hbar}({\alpha p + \beta x})}\left|x\right>$$ $$=e^{-{i\over 2\hbar}\alpha\beta}\int dx \left<x\right|e^{{i\over \hbar}{\beta x}}e^{{i\over \hbar}{\alpha p}}\left|x\right>$$ by the BCH formula. But the 2 in the exponent's denominator seems to do not match with the result.

  2. How can I construct $N$-dimensional basis from them, and

  3. Is there any other way to construct $N$-dimensional unitary basis?

Any kind of help will be appreciated!

  • 2
    $\begingroup$ 1. The operators $x$ and $p$ exist only on infinite-dimensional spaces, why do you think you can construct a finite-dimensional unitary basis from them? 2. The operator $E(\alpha,\beta)$ is most probably not trace-class, since neither $x$ nor $p$ are, and writing that a trace is a delta function is rigorously non-sensical. In what sense do you think you can take the trace here? $\endgroup$ – ACuriousMind Aug 2 '15 at 12:27
  • $\begingroup$ @ACuriousMind I have never been taught about (even basic) operator theory, so I was blind to that, but now I see your point. But by following the very formal definition of the trace of the operator with the continuous spectrum, I end up with the similiar form of the Fourier transform of the exponential function (which is the delta function), and I think that's what Wheeler meant on his note. Dosen't the fact that the trace is not a scalar but the delta function tell us that we can define 'orthonormal basis' in some sense? $\endgroup$ – generic properties Aug 2 '15 at 14:31
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    $\begingroup$ You are basically looking for representations of the Lie algebra of $U(N)$. The easiest way to construct a representation is to take the adjoint representation of $SU(N)$, and add the identity matrix which accounts for the additional $U(1)$ factor. $\endgroup$ – Meng Cheng Aug 2 '15 at 15:40

(1). The degenerate CBH formula is meant to clear the underbrush for you: $$\text{Tr}(E(\alpha,\beta)) = \int dx<x|e^{{i\over \hbar}({\alpha p + \beta x})}|x>\\ =e^{-{i\over 2\hbar}\alpha\beta}\int dx <x|e^{{i\over \hbar}{\beta x}}e^{{i\over \hbar}{\alpha p}}|x>= e^{-{i\over 2\hbar}\alpha\beta}\int dx ~e^{{i\over \hbar}{\beta x}}<x|e^{{i\over \hbar}{\alpha p}}|x> \\ =e^{-{i\over 2\hbar}\alpha\beta}\int dx ~e^{{i\over \hbar}{\beta x}}<x| x+ \alpha >= e^{-{i\over 2\hbar}\alpha\beta} \delta(\alpha)\int dx ~e^{{i\over \hbar}{\beta x}}\\ = e^{-{i\over 2\hbar}\alpha\beta} ~\delta(\alpha) ~\delta \left(\frac{\beta}{2\pi \hbar}\right)=h \delta(\alpha) \delta (\beta), $$ where you use the action of the translation operator, the Fourier representation of the delta function, the scaling property of it, and collapse the exponent of the term you were baffled about at its argument.

This is the large $N$ generalization of the celebrated trace-orthogonal unitary clock-and-shift -matrix basis of U(N) utilized by Weyl in the context of QM in finite Hilbert spaces--it was actually invented in 1867 by J. J. Sylvester. $2\pi/N\leftrightarrow \hbar$.

You may explain the construction to yourself for N =3, Sylvester's "nonions" , constructed almost a century before Gell-Mann's su(3) Lie algebra matrices. (Clock, $Q\sim V\sim \Sigma_3$, shift, $P\sim U\sim \Sigma_1$, in evocation of generalizing Pauli matrices. Needless to say, I'd supplant "others" with "Sylvester" in your title. References to Schwinger, Werner, etc... are parochial and laughable.)

(2). As the commentators point out, the Heisenberg Lie algebra has no finite dimensional unitary representations, but you might go the other way, $N\to \infty$, e.g. in Q10 of this exam; actually, the original Santhanam & Tekumalla paper is not that hard. In the finite N case, α and β take discrete values on a 2-torus, and only become continuous for large N. Now, 1/h ~ N, that is, the Kronecker δs of the N×N identity matrix devolve to the δ-functions for large N. But you can't get by with finite N.

(3) I don't think so. You might be tempted by the mirage of the formal Heisenberg group $H_3(R)$, $$ \begin{bmatrix} 1 & \beta & c\\ 0 & 1 & \alpha \\ 0 & 0 & 1\\ \end{bmatrix}= \exp \begin{bmatrix} 0 & \beta & c-\alpha\beta/2\\ 0 & 0 & \alpha \\ 0 & 0 & 0\\ \end{bmatrix} , $$ but note the central element in the Heisenberg algebra here, gotten from the logarithm at α=β=0, is not the identity! The trace of these upper triangular elements is always the identity, and this is evidently not a unitary basis.


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